Let \(H_1\le\mathbb Z^2\) be generated by \((1,2)\) and \((4,1)\), and let \(H_2\le\mathbb Z^2\) be generated by \((3,2)\) and \((1,3)\). Are the quotient groups \[ \mathbb Z^2/H_1 \qquad\text{and}\qquad \mathbb Z^2/H_2 \] isomorphic?
Solution
Let \(e_1=(1,0)\) and \(e_2=(0,1)\).
1. The quotient \(\ZZ^2/H_1\) is generated by the class of \(e_2\), andthat class has order dividing \(7\).
In \(\ZZ^2/H_1\), the generator \((1,2)=e_1+2e_2\) of \(H_1\) gives \[ [e_1]=-2[e_2]. \] Hence every class is a multiple of \([e_2]\). The second generator \((4,1)=4e_1+e_2\) then gives \[ 0 = 4[e_1]+[e_2] = -8[e_2]+[e_2] = -7[e_2]. \] Thus \(7[e_2]=0\).
2. In fact,\[ \ZZ^2/H_1\cong\ZZ/7\ZZ. \]
Define \[ \phi_1:\ZZ^2\to\ZZ/7\ZZ, \qquad \phi_1(m,n)=5m+n\pmod7. \] Then \[ \phi_1(1,2)=5+2=0 \qquad\text{and}\qquad \phi_1(4,1)=20+1=0 \] in \(\ZZ/7\ZZ\). Thus \(H_1\subseteq\ker\phi_1\), so \(\phi_1\) induces \[ \overline{\phi}_1:\ZZ^2/H_1\to\ZZ/7\ZZ. \] Since \[ \overline{\phi}_1([e_2])=1, \] this induced map is surjective. Step 1 shows that its domain is cyclic generated by an element of order dividing \(7\), while surjectivity forces that generator to have order at least \(7\). Hence its order is exactly \(7\), and the induced map is an isomorphism.
3. The quotient \(\ZZ^2/H_2\) is generated by the class of \(e_2\), andthat class has order dividing \(7\).
In \(\ZZ^2/H_2\), the relation \[ e_1+3e_2=0 \] from the generator \((1,3)\) gives \[ [e_1]=-3[e_2]. \] Thus \([e_2]\) generates the quotient. The other generator \((3,2)=3e_1+2e_2\) gives \[ 0 = 3[e_1]+2[e_2] = -9[e_2]+2[e_2] = -7[e_2]. \] Therefore \(7[e_2]=0\).
4. In fact,\[ \ZZ^2/H_2\cong\ZZ/7\ZZ. \]
Define \[ \phi_2:\ZZ^2\to\ZZ/7\ZZ, \qquad \phi_2(m,n)=4m+n\pmod7. \] Then \[ \phi_2(3,2)=12+2=0 \qquad\text{and}\qquad \phi_2(1,3)=4+3=0 \] in \(\ZZ/7\ZZ\). Hence \(H_2\subseteq\ker\phi_2\), and \(\phi_2\) induces a surjection \[ \overline{\phi}_2:\ZZ^2/H_2\to\ZZ/7\ZZ \] because \(\overline{\phi}_2([e_2])=1\). Step 3 now shows, exactly as in step 2, that \([e_2]\) has order exactly \(7\) and the induced map is an isomorphism.
5. Therefore\[ \boxed{\ZZ^2/H_1\cong\ZZ^2/H_2}. \]
By steps 2 and 4, both groups are isomorphic to \(\ZZ/7\ZZ\).
6. Q.E.D.
Step 5 answers the question.